If \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{k}{a+b+c+d}\ge \frac{16+k}{4}\] holds for all $a,b,c,d>0,abcd=1$ ,then$k_{max}=k_0=12.4185242..., $ where \[-1938817024-334692352k_0-172032k_0^2+2304512k_0^3+62768k_0^4+729k_0^5=0.\]

It's CGMO2011,p3 I have proved this inequality in a way which is not so nice……

Try also this one:(my solution work on this too) \[abcd = 1,a,b,c,d > 0\] \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{{16}}{{a + b + c + d}} \ge \frac{{40\sqrt 3 }}{9}\] When does the equality hold?

Show us your slution, please!

I'll prove a stronger inequality: Let $a$, $b$, $c$ and $d$ are positives such that $abcd=1$. Prove that: \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7\] Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$. Hence, $f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$ $=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(a+b+c+d)\left(a+3\sqrt[3]{bcd}\right)}\geq$ $\geq\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(\frac{b+c+d}{3}+b+c+d)\left(\frac{b+c+d}{3}+3\sqrt[3]{bcd}\right)}=$ $=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}$. We'll prove that $\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0$. Let $b+c+d=3u$, $bc+bd+cd=3v^2$ and $bcd=w^3$. Hence, $\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0\Leftrightarrow g(v^2)\geq0$, where $g$ is linear increasing function. Hence, $g$ gets a minimal value, when $v^2$ gets a minimal value, which happens, when two numbers from $\{b,c,d\}$ are equal. Since $g(v^2)\geq0$ is homogeneous, for the proof of $g(v^2)\geq0$ we need to check only one case: $c=d=1$, which after substitution $b=x^3$ gives $(x-1)^2(2x^7+x^6+18x^5-10x^4-50x^3+36x^2+26x+4)\geq0$, which is true. Id est, $f(a,b,c,d)\geq f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)$ and it remains to prove that $f(a,b,b,b)\geq0$, where $a=\frac{1}{b^3}$, which gives $(b-1)^2(3b^6+6b^5+9b^4-9b^3-5b^2-b+3)\geq0$, which is true. By the same way we can check the fjwxcsl's inequality.

I really want to see a solution without MV method...My solution is not compeletely MV but a bit like that

What is CGMO? Is official solution also by MV method?

mudok wrote: What is CGMO? Is official solution also by MV method? It is China Girls Math Olympiad I havent seen the official solution yet……

red3 wrote: The positive reals $a,b,c,d$ satisfy $abcd=1$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$. See here: http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=96&id=23073 fjwxcsl wrote: If \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{k}{a+b+c+d}\ge \frac{16+k}{4}\] holds for all $a,b,c,d>0,abcd=1$ ,then$k_{max}=k_0=12.4185242..., $ where \[-1938817024-334692352k_0-172032k_0^2+2304512k_0^3+62768k_0^4+729k_0^5=0.\]. fjwxcsl wrote: Determine the maximum $k_1$ such that $(\forall{x_1,x_2,x_3,x_4}>0)x_{1}+x_{2}+x_{3}+x_{4}+\frac{4k_1}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}} \ge{(4+k_1){\sqrt[4]{x_{1}x_{2}x_{3}x_{4}}}}.$ FROM《fjwxcsl:210 nice symmetric inequalities(4,5),$T_{4.9(1)}$》,SEE: http://www.irgoc.org/viewtopic.php?f=10&t=926&sid=59fe05e18960535ca0c6b19a0394321c&start=0

but what if k>16?

red3 wrote: but what if k>16? The answer is very ugly

The positive reals $a,b,c,d,e$ satisfy $abcde=1$. Prove that \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{16}{{a + b + c + d+e}} \ge{\frac{{41}}{5}}.\]

Here is a similar result, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=354052 Tourish wrote: $x_1,x_2,\cdots,x_n>0$,prove that \[x_1+x_2+\cdots+x_n+\frac{n(n-1)}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}}\geq k_n\sqrt[n]{x_1x_2\cdots x_n}\] where \[k_n= \begin{cases} 2n-1,&2\leq n\leq 4\\ \sqrt[n]{\frac{2(n-1)}{n-2-\sqrt{n(n-4)}}}\cdot\frac{2n\left[n^2-2n-(n-1)\sqrt{n(n-4)}\right]}{(n-1)\left[n-\sqrt{n(n-4)}\right]},&n\geq 5 \end{cases} \] where $k_n$ has the best result .

just by $EV$,one can get it.

if one doesn't know about $EV$ theorem,Lagrange is still available. Let us consider the Lagrange function $\sum_{cyc}abc+\frac{9}{\sum_{ }a}+t(abcd-1)$. by easy derivative calculations and substitutions we get $a=b$ or $cd(\sum_{cyc}a)^2=9$; $a=c$ or $bd(\sum_{cyc}a)^2=9$: $a=d$ or $bc(\sum_{cyc}a)^2=9$: $abcd=1$. by some discussions it's trivial that $(a,b,c,d)=(x,x,x,\frac{1}{x^3})$or$(x,x,\frac{1}{x^2},\frac{1}{x^2})$ this coincides with the results obtained by $EV$.

Can anyone prove the original inequality in more basic way?

yes,but it's very difficult.the basic way is adjustment.

such ineqs are always ready to be second-killed!you can solve them by any possible method you'd like to use.the truely difficult ones are ineqs with n variables,especially if they can't be proved by induction~

$\frac{1}{a^4} + \frac{1}{b^4} + \frac{1}{c^4} + \frac{1}{d^4} + \frac{9}{a^4 + b^4 + c^4 + d^4} $ $\geq\frac{8}{9}\left(\frac{1}{a^2b^2}+\frac{1}{a^2c^2}+\frac{1}{a^2d^2}+\frac{1}{b^2c^2}+\frac{1}{b^2d^2}+\frac{1}{c^2d^2}\right)+ \frac{11}{3\left(a^4 + b^4 + c^4 + d^4\right)}$ $\geq\frac{25}{4abcd},$ where the last inequality follows from $36a^2b^2c^2d^2\left(a^4 + b^4 + c^4 + d^4\right)\left[\frac{11}{3\left(a^4 + b^4 + c^4 + d^4\right)}-\frac{25}{4abcd}+\frac{8}{9}\sum_{sym}{\frac{1}{a^2b^2}\right]}$ $=\sum_{sym}{\frac{(a-b)^2}{12}\Big\{(a-b)^2\left[192\left(a^2+b^2\right)\left(d^2+c^2\right)+ab\left(331c^2-186cd+331d^2\right)+80cd(c+d)(a+b)+80c^2d^2\right]}$ ${+3cd(c+d)^2\left(18ab+119c^2-194cd+119d^2\right)\Big\}}\geq 0,$ and the first inequality see: http://www.mathoe.com/dispbbs.asp?boardID=107&ID=44556

robinpark wrote: Consider the functions $f(a, b, c, d) = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}$ and $g(a, b, c, d) = abcd-1$. By the Method of Lagrange Multipliers, we have $\nabla f = \lambda \nabla g.$ Hence we have $\frac{\partial f}{\partial a} = \lambda \frac{\partial g}{\partial a}$, or\[-\frac{1}{a^2}-\frac{9}{(a+b+c+d)^2} = \lambda(bcd)\]from which we obtain $\lambda = -\frac{1}{a}-\frac{9a}{(a+b+c+d)^2}$. Similarly, $\lambda = -\frac{1}{b}-\frac{9b}{(a+b+c+d)^2}$, etc. Setting two of these values equation to each other, we have $\frac{1}{a}+\frac{9a}{(a+b+c+d)^2} = \frac{1}{b}+\frac{9b}{(a+b+c+d)^2}$, implying that $(a+b+c+d)^2 = 9ab$ after factoring out $b-a$. Since this is true symmetrically, we have that $9ab = 9bc = 9cd = 9da = 9ac = 9bd$, implying that $a=b=c=d$. Setting this into $f$, we see that $f(1, 1, 1, 1) = \frac{25}{4}$ is the extremum, and it is not difficult to see that values greater than $\frac{25}{4}$ are attainable (for example, $f\left(1, 1, 2, \frac{1}{2}\right) = \frac{13}{2} > \frac{25}{4}$), so $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d} \ge \frac{25}{4}$. Very nice solution Is this correct too?! Since:$\frac{\lambda}{a}+\frac{1}{a^2}=\frac{\lambda}{b}+\frac{1}{b^2}=\frac{\lambda}{c}+\frac{1}{c^2}=\frac{\lambda}{d}+\frac{1}{d^2}$ So we get six equations like: $(\frac{1}{a}-\frac{1}{b})(\lambda+\frac{1}{a}+\frac{1}{b})=0$ And from these we get that $a=b=c=d$

With out loss of generality we may assume $ab\geq cd$. Let $ab=x^2$, $cd=1/x^2$, then $x\geq1$. Set $u=a+b+c+d$, $v=c+d$, then we have $u\geq 2x+2/x$, $v\geq 2/x$, and the left hand side of the inequality becomes $$ \frac{u}{x^2} + v(x^2-\frac{1}{x^2}) + \frac{9}{u} \geq \frac{u}{x^2} + \frac{9}{u} + \frac{2}{x}(x^2 - \frac{1}{x^2}).$$ For each fixed $x$, then function $f(u) = \frac{u}{x^2} + \frac{9}{u}$ reached its minimum at $u=3x$. But in the present situation, $u\geq 2x+2/x$, hence we may discuss (1) If $3x \geq 2x + 2/x$, namely, $x\geq\sqrt{2}$, then we have $f(u)\geq 6/x$ and $$ LHS \geq \frac{6}{x} + \frac{2}{x}(x^2 - \frac{1}{x^2}) \geq \frac{9}{\sqrt{2}} > \frac{25}{4}.$$ (2) If $3x < 2x + 2/x$, namely, $1\leq x < \sqrt{2}$, then we have $f(u) \geq f(2x + 2/x)$ and $$ LHS \geq \frac{2x+2/x}{x^2} + \frac{9}{2x+2/x} + \frac{2}{x}(x^2 - \frac{1}{x^2}) = (2x+2/x)+\frac{9}{2x+2/x} \geq\frac{25}{4}.$$

Kunihiko_Chikaya wrote: red3 wrote: The positive reals $a,b,c,d$ satisfy $abcd=1$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$. I have just found the geometric proof as follows. Regrettably, the proof was not perfect, but I dare to remain as memorial. Kunihiko Chikaya Take points $A\left(a,\ \frac{1}{a}\right),\ B\left(b,\ \frac{1}{b}\right),\ C\left(c,\ \frac{1}{c}\right),\ D\left(d,\ \frac{1}{d}\right)$ on the hyperbola $y=\frac{1}{x}\ (x>0),$ W.L.O.G. $0<a\leq b\leq c\leq d$. Let $K,\ L$ be the midpoints of the line segments $AB,\ CD$ respectively, we have $K\left(\frac{a+b}{2},\ \frac{\frac{1}{a}+\frac{1}{b}}{2}\right),\ L\left(\frac{c+d}{2},\ \frac{\frac{1}{c}+\frac{1}{d}}{2}\right)$, we have the midpoint $M\left(\frac{a+b+c+d}{4},\ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4}\right)$ of the line segment $KL$. Let $N\left(\frac{a+b+c+d}{4},\ \frac{1}{\frac{a+b+c+d}{4}}\right).$ Since $y=\frac{1}{x}\ (x>0)$ is a convex and decreasing function, take a point $P$ such that $MP:PN=9:16$ on the line segment $MN$, denoting by $y_{P}$ the $y$-coordinate of the point $P$, we have $y_{N}\leq y_{P}\leq y_{M}\ \cdots (*)$, then by the AM-GM inequality, from the condition $abcd=1$, we have $y_N=\frac{1}{\frac{a+b+c+d}{4}}\leq \frac{1}{\sqrt[4]{abcd}}$, yielding $y_N\leq 1.$ Therefore from $(*)$, we have $y_P=\frac{16\cdot \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}{4}+9\cdot \frac{1}{\frac{a+b+c+d}{4}}}{9+16}\geq 1$ $\Longleftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\geq \frac{25}{4}.$ Equality holds when $a=b=c=d=1$, or all of points $A,\ B,\ C,\ D$ coincide. $y_p\leq 1$ and $y_p\leq y_n$ does not imply $y_n\geq 1$, sir.

red3 wrote: The positive reals $a,b,c,d$ satisfy $abcd=1$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$. I tried to find an elementary proof. Used only Schur, AM-GM and factorizations. After substitutions $a\to \frac{1}{a}, ... $ we need to prove $a+b+c+d+\frac{9}{abc+bcd+cda+dab}\ge \frac{25}{4}$ A weak lemma: $(a+b+c+d)^3+\frac{128abcd}{a+b+c+d}\ge 24(abc+bcd+cda+dab)$ Proof of the lemma: WLOG $a+b+c+d=4, \ \ d=min(a,b,c,d)\le 1$. We need to prove $32+16abcd\ge 12abc+3d\cdot 4(ab+bc+ca)$ $\Leftarrow 32+16abcd\ge 12abc+3d ((a+b+c)^2+\frac{9abc}{a+b+c}) \ \ \ \ \ \ \text{(by Schur's inequality)}$ $\Leftarrow 32\ge 12abc(1-d)+3d(a+b+c)^2+ abc( \frac{27}{a+b+c}-4)d \ \ \ \ \ \text{By AM-GM:}$ $\Leftarrow 32\ge \frac{4(a+b+c)^3}{9} (1-d)+3d(a+b+c)^2+ \frac{(a+b+c)^3}{27} \cdot( \frac{27}{a+b+c}-4)d$ $\Leftarrow 32\ge \frac{4(4-d)^3}{9} (1-d)+3d(4-d)^2+ \frac{(4-d)^3}{27} \cdot( \frac{27}{4-d}-4)d$ $(d-1)^2(6+4d-d^2)\ge 0$. Proof of the lemma is finished. Let $a+b+c+d=x\ge 4$. By the lemma we have $\frac{x^4+128}{24x}\ge abc+bcd+cda+dab$. So it is enough to prove $x+\frac{9\cdot 24x}{x^4+128}\ge \frac{25}{4} \ \ \ \Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 800)\ge 0$ $\Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 704)\ge 0$ $\Leftarrow \ \ (x - 4)^2 (4 x^3 + 7 x^2 - 8 x - 176)\ge 0$ Which is obvious due to $x\ge 4$.

arqady wrote: Let $a$, $b$, $c$ and $d$ are positives such that $abcd=1$. Prove that: \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7\] Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$$

sqing wrote: Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$$ Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{n^2(\sqrt[n]{2n}-1)}{a_1+a_2+\cdots+a_n}\geqslant n\sqrt[n]{2n}. $$(Jichen) p/4089769260

mudok wrote: red3 wrote: The positive reals $a,b,c,d$ satisfy $abcd=1$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} \geqslant \frac{{25}}{4}$. I tried to find an elementary proof. Used only Schur, AM-GM and factorizations. After substitutions $a\to \frac{1}{a}, ... $ we need to prove $a+b+c+d+\frac{9}{abc+bcd+cda+dab}\ge \frac{25}{4}$ A weak lemma: $(a+b+c+d)^3+\frac{128abcd}{a+b+c+d}\ge 24(abc+bcd+cda+dab)$ Proof of the lemma: WLOG $a+b+c+d=4, \ \ d=min(a,b,c,d)\le 1$. We need to prove $32+16abcd\ge 12abc+3d\cdot 4(ab+bc+ca)$ $\Leftarrow 32+16abcd\ge 12abc+3d ((a+b+c)^2+\frac{9abc}{a+b+c}) \ \ \ \ \ \ \text{(by Schur's inequality)}$ $\Leftarrow 32\ge 12abc(1-d)+3d(a+b+c)^2+ abc( \frac{27}{a+b+c}-4)d \ \ \ \ \ \text{By AM-GM:}$ $\Leftarrow 32\ge \frac{4(a+b+c)^3}{9} (1-d)+3d(a+b+c)^2+ \frac{(a+b+c)^3}{27} \cdot( \frac{27}{a+b+c}-4)d$ $\Leftarrow 32\ge \frac{4(4-d)^3}{9} (1-d)+3d(4-d)^2+ \frac{(4-d)^3}{27} \cdot( \frac{27}{4-d}-4)d$ $(d-1)^2(6+4d-d^2)\ge 0$. Proof of the lemma is finished. Let $a+b+c+d=x\ge 4$. By the lemma we have $\frac{x^4+128}{24x}\ge abc+bcd+cda+dab$. So it is enough to prove $x+\frac{9\cdot 24x}{x^4+128}\ge \frac{25}{4} \ \ \ \Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 800)\ge 0$ $\Leftarrow \ \ (x - 4) (4 x^4 - 9 x^3 - 36 x^2 - 144 x + 704)\ge 0$ $\Leftarrow \ \ (x - 4)^2 (4 x^3 + 7 x^2 - 8 x - 176)\ge 0$ Which is obvious due to $x\ge 4$. beautifull prove

sqing wrote: arqady wrote: Let $a$, $b$, $c$ and $d$ are positives such that $abcd=1$. Prove that: \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}\ge 7\] Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$$ https://artofproblemsolving.com/community/c6h1810691p12061378 Let $a,b,c,d>0$ such that $abcd=1$. prove that: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{289}{a+b+c+d+23} \ge \frac{397}{27}$$

WLOG $ab \ge cd$ $\quad$ $\therefore ab \ge 1$ $a+b \ge 2\sqrt{ab}, c+d \ge 2\sqrt{cd}=\frac{2}{\sqrt{ab}}$, Assume $\sqrt{ab}=x$ $a+b+c+d \ge 2 (x+\frac{1}{x})$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{{a + b + c + d}} = \frac{a+b+c+d}{ab}+(c+d)(ab-cd)+\frac{9}{{a + b + c + d}}=\frac{k}{x^2}+\frac{9}{k}+(c+d)(x^2-\frac{1}{x^2})$ where $k=a+b+c+d$ Case 1: $3x \ge 2(x+\frac{1}{x}) \implies x \ge \sqrt{2}$ Lemma 1: $\frac{k}{x^2}+\frac{9}{k} \ge \frac{6}{x}$ Proof of lemma 1: $\frac{k}{x^2}+\frac{9}{k} \ge \frac{6}{x} \iff k^2 + 9x^2 \ge 6xk \iff (k-3x)^2 \ge 0$ which is obviously true. $\blacksquare$ $ LHS \ge \frac{6}{x}+(c+d)(x^2-\frac{1}{x^2}) \ge \frac{6}{x}+ \frac{2}{x}(x^2-\frac{1}{x^2})$ Assume $f(x)=\frac{6}{x}+ \frac{2}{x}(x^2-\frac{1}{x^2})=6x^{-1}+2x-2x^{-3}$ So, $f'(x)=\frac{2}{x^4}(x^4-3x^2+3)=\frac{2}{x^4}((x^2-1.5)^2+0.75)$ which is always positive when $x \ge \sqrt{2}$. which means $f(x)$ is an increasing function. So the least value can be achieved when $x = \sqrt{2}$ $\therefore LHS \ge f(\sqrt{2})=\frac{9}{\sqrt2} \ge \frac{25}{4}$ Case 2: $3x < 2(x+\frac{1}{x}) \le k$ Assume $g(k)=\frac{k}{x^2}+\frac{9}{k}$ So, $g'(k)=\frac{1}{x^2}-\frac{9}{k^2} = \frac{k^2-9x^2}{x^2k^2}$ which is positive since $k>3x$ This implies $g(k)$ is an increasing function when $k>3x$ $\therefore LHS \ge g(k)+\frac{2}{x}(x^2-\frac{1}{x^2}) \ge g(2(x+\frac{1}{x}))+\frac{2}{x}(x^2-\frac{1}{x^2})=\frac{2(x+\frac{1}{x})}{x^2}+\frac{9}{2(x+\frac{1}{x})}+\frac{2}{x}(x^2-\frac{1}{x^2})=(2x+\frac{2}{x})+\frac{9}{2x+\frac{2}{x}}=y+\frac{9}{y}$ where $y=2x+\frac{2}{x} \ge 4$ Assume $h(y)=y+\frac{9}{y}$ So, $h'(y)=1-\frac{9}{y^2}=\frac{y^2-9}{y^2}$, since $y \ge 4 >3$, this derivative must be positive, so $h(y)$ is an increasing function too. Thus, $LHS \ge h(y) \ge h(4) = \frac{25}{4}$ Q.E.D

oldbeginner wrote: Show us your slution, please! If it's still needed,just contact me in PM and I'll send my solution. At this moment,I will not post it publicly.

1/a+1/b+1/c+1/d>=16/a+b+c+d. a+b+c+d>=4. 1/a+1/b+1/c+1/d+ 9/a+b+c+d>=25/4

Say...an wrote: 1/a+1/b+1/c+1/d>=16/a+b+c+d. a+b+c+d>=4. 1/a+1/b+1/c+1/d+ 9/a+b+c+d>=25/4 How does the first one implies the second. thats totally unrelated. And moreover, $\frac{16}{a + b + c + d} \le 4 $.

I also wrote it...in different way ....see it carefully

sqing wrote: Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{4(n-1)}{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+4n-4}}{n}$$ Let $a_1,a_2,\cdots,a_n$ are positives such that $a_1a_2\cdots a_n=1$ $(n\ge 2)$. Prove that $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}+\frac{\lambda }{a_1+a_2+\cdots+a_n}\geqslant \frac{{n^2+\lambda }}{n}$$Where $0\le \lambda \le4(n-1).$ (SXTX (3)2020)

The positive reals $a,b,c,d$ satisfy $3(a^2+b^2+c^2+d^2)=4(a+b+c+d)+1$. Prove that $$a^3+b^3+c^3+d^3\leq11.$$Where?